Answer
$$\lim_{x \to 0}f(x)=0$$
Work Step by Step
Looking at the graph, we find that the function $f(x)=x \sin \frac{1}{x}$ is squeezed between $|x|$ and $-|x|$, $-|x| \le x \sin \frac{1}{x} \le |x|$, over some interval containing $x=0$. We also find that$$\lim_{x \to 0}-|x|= \lim_{x \to 0}|x|=0 .$$Thus, by applying the Squeeze Theorem we conclude that$$\lim_{x \to 0}f(x)= \lim_{x \to 0}x \sin \frac{1}{x}=0.$$
